今年秋招 JPMorgan Chase 的 OA 已經出了，題型延續了一貫的風格：思維邏輯+編程實現，不算特別“花哨”，但如果沒刷到類似題目，很容易在邊界條件上卡殼。下面給大家分享兩道真題，都是我們學員第一時間反饋的。
JPMorgan Chase 招聘 / 投遞流程（官方 & 經驗匯總）
官方流程與政策（JPMorgan Careers “How We Hire”）
- 在 JPM Chase 的官網 “How We Hire” 頁面可以看到其招聘流程框架：提交申請 → 面試 → 錄用。
- 在 FAQ 部分，有關於申請狀態、撤回申請、草稿保存、是否可以更改已提交申請的信息。
- 如果某個崗位表示“需要評估”的（Assessments / 在線測評），你會在申請流程中被通知是否需要做測評。
Signal Filtering
Signal filtering based on frequency is essential to minimize noise outside the desired frequency range. Filters can be combined such that onlyfrequencies within the permissible range of all filters can pass through.
For instance, three filters with frequency ranges (10, 17), (13, 15), and (13, 17) will only allow frequencies between 13 and 15 to pass through, as this isthe only overlapping range.
Given n signal frequencies and a sequence of m filters with specified frequency ranges from x to y (inclusive), determine how many signals will passthrough all the filters. There will be a single common range where all filters overlap.
Example:
frequencies = [8, 15, 14, 16]
filtersRanges = [[10, 17], [13, …]]
The range that all of the filters overlap is from … to 15, inclusive. The 2 frequencies that will pass through
思路：
典型的區間交集問題。所有 filter 取一個最大下界 L = max(x) 和最小上界 R = min(y)，最終只保留 [L, R] 區間內的頻率。遍歷一遍 frequencies 統計就行。時間複雜度 O(n+m)。
Two Consignments
A shop has n types of items, where the quantity of the i-th item is denoted by quantity[i]. These items are to be shipped in two consignments:
- The first consignment contains items 1, 2, …, j
- The second consignment contains the remaining items j+1, …, n
The value of j can be chosen such that 1 ≤ j < n (1-based indexing).
The shopkeeper wants to make the total quantities in both consignments equal. To achieve this, they can increase or decrease the quantity of any item typeby 1, any number of times. However, the quantity of each item must remain positive.Find the minimum number of operations required to make the quantities of both consignments equal if the items are split optimally.
Example:
n = 3
quantity = [1, 4, 4]
The optimal approach is to increase quantity = [1, 4, 5].
Partition at j = 2 to get consignments [1, 4] and [5].
The answer is 1.
思路：
這道題看似複雜，其實就是在所有分割點 j 中找一個最優的。對於每個分割點，左邊和右邊的總和分別是 A 和 B。如果兩邊其中一側的總和足夠大（≥ 類別數），只要調整差值就能搞定，操作數就是 |A-B|。
但如果兩邊的和都太小（小於該側的類別數），就必須整體抬升到最低要求，這時公式會變成 2*Lmin - (A+B)。枚舉所有 j，取最小值即可。時間複雜度 O(n)。
總結
這兩道題一個是區間交集、一個是數組分割平衡，看似獨立，其實都考察了邊界條件的處理能力：
- Filter 那題如果不注意 inclusive（包含端點），很容易算錯。
- Consignment 那題如果忽略了每類數量必須 ≥1，就會漏掉特殊情況。
很多同學平時刷題思路沒問題，但一旦在 OA 里卡到這種邊界，就容易超時或寫崩。
不只是刷題，真正的底氣來自這裡
我們 programhelp 团队在这类大厂 OA 上有非常多的实战经验：
- 远程无痕辅助：幫你寫出 100% 過測的代碼，不留痕跡；
- 實時語音助攻：你在做題卡殼時，我們能及時提醒你該注意的邊界點；
- VO/面試全流程輔導：不僅限於 OA，還能幫你穩住後續技術面。
已經有不少學員靠著我們的助攻順利通過 JPMorgan 的 OA，進入後續面試環節。
如果你也在準備今年的秋招，不想在 OA 上栽跟頭，可以隨時聯繫我們 。
